HA9 Problem 3

[Victor Ivrii]

Using method of reflection (studied earlier for different equations) construct Green function for

1.  Dirichlet problem
2.  Neumann problem

for Laplace equation in

1.  half-plane
2.  half-space

as we know that in the whole plane and space they are just potentials
\begin{gather}
\frac{1}{2\pi}\log \bigl((x_1-y_1)^2+(x_2-y_2)^2\bigr)^{\frac{1}{2}},\label{equ-H9.3}\\
-\frac{1}{4\pi}
\bigl((x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2\bigr)^{-\frac{1}{2}}
\label{equ-H9.4}
\end{gather}
respectively.

[Chaojie Li]

Q3(a)
Dir Problem in the hall plane for Laplace equation :
In whole place $G(x,y)=\frac{1}{2\pi}\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} $ where $x=(x_1,x_2), y=(y_1,y_2)$
consider the upper half plane $y>0$ yes odd continuation
$$-y=(y_1,-y_2)\implies f(-y)=-f(y)$$
so
$$u(x,y)=\int_0^\infty f(x,y)G(x-y)dy+\int_{-\infty}^0f(x-y)G(x-y)dy$$
$$=\int_0^\infty f(x,y)G(x-y)dy+\int_0^\infty-f(x,y)G(x+y)dy$$
$$=\int_0^\infty f(x,y)[G(x-y)-G(x+y)]dy$$
$$G(x-y)-G(x+y)=\frac{1}{2\pi}\log \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}-\frac{1}{2\pi}$$
$$\frac{1}{2\pi} \log \sqrt{\frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}}$$
$$=\frac{1}{4\pi}\log \frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}$$
For upper half plane $y>0$, Green function $G(x,y)=\frac{1}{4\pi}\log \frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}$
similarly Dirichlet problem in half space for Laplace equation
$$G=-\frac{1}{4\pi}[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2]^{1\frac{1}{2}}$$
where $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$,$-y=(y_1,y_2,-y_3)$
for $z>0$ upper half space:
$$G(x,y)=G(x-y)-G(x+y)=-\frac{1}{4\pi}\{[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_2)^2]^{-\frac{1}{2}}-[(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2]^{-\frac{1}{2}}\}$$
Green function. $\blacksquare$

[Chaojie Li]

(b)
As in part(a) we can do similar by reflection for Neumann problem for Laplace equation in brief use even continuation $f(-y)=f(y),x=(x_1,x_2),y=(y_1,y_2),-y=(y_1,-y_2)$\\\
Upper half plane, $y>0$
$$u(x,y)=\int_0^\infty f(x,y)G(x-y)dy+\int_0^\infty f(x,-y)G(x-y)dy$$
$$=\int_0^\infty f(x,y)G(x-y)dy +\int _0^\infty f(x,y)G(x+y)dy$$
$$=\int _0 ^\infty f(x,y)[G(x-y)+G(x+y)]dy$$
$$G(x-y)-G(x+y)=\frac{1}{2\pi}\log [(x_1-y_1)^2+(x_2-y_2^2)]^{\frac{1}{2}}+\frac{1}{2\pi}\log[(x_1-y_1)^2+(x_2+y_2)^2]^{\frac{1}{2}}$$
$$=\frac{1}{4\pi}\log [(x_1-y_1)^2+(x_2-y_2)^2][(x_1-y_1)^2+(x_2+y_)^2]$$
For upper half plane $y>0$, green function:
$$G(x,y)=\frac{1}{4\pi}\log [(x_1-y_1)^2+(x_2-y_2)^2][(x_1-y_1)^2+(x_2+y_2)^2]$$
Similarly, we consider the upper half space $z>0$,$x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$,$-y=(y_1,y_2,-y_3)$\\
$$G(x,y)=G(x-y)+G(x+y)=-\frac{1}{4\pi}\{[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2]^{-\frac{1}{2}}+[(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2]^{-\frac{1}{2\pi}}\}$$
for $z>0$ upper half space , Green Function
$$G(x,y)=-\frac{1}{4\pi \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}}-\frac{1}{4\pi \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2}}$$

[Mark Nunez]

A slightly different derivation.