HA9 Problem 2

[Victor Ivrii]

Apply method of descent described in [Subsection 26.4](./L26.html#sect-26.4) but to Laplace equation in $\mathbb{R}^2$ and starting from Coulomb potential in $3D$
\begin{equation}
U_3(x,y,z)=-\frac{1}{4\pi} \bigl(x^2+y^2+z^2\bigr)^{-\frac{1}{2}},
\label{equ-H9.1}
\end{equation}
derive logarithmic potential in $2D$
\begin{equation}
U_2(x,y,z)=\frac{1}{2\pi}\log \bigl(x^2+y^2\bigr)^{\frac{1}{2}},
\label{equ-H9.2}
\end{equation}
*Hint.* You will need to calculate diverging integral $\int_0^\infty U_3 (x,y,z)$. Instead consider $\int_0^N U_3 (x,y,z)\,dz$, subtract constant (f.e. $\int_0^N U_3 (1,0,z)\,dz$) and then tend $N\to \infty$.

[Chaojie Li]

$$u(x,y,z)=\int U_3(x-x',y-y',z-z')f(x',y',z')dx'dy'dz'$$
Since $z-z'$ is a constant so that u doesn't depend on $z'$
$$u(x,y)=\int U_3(x-x',y-y',z-z')dZdx'dy' =\int U_2 (x-x',y-y') dx'dy'$$
so $U_2=\int U_3(x-x',y-y',z-z')dZ$\\
and we have :
$$U_3=-\frac{1}{4\pi}(x^2+y^2+z^2)^{\frac{1}{2}}$$
$$\text{and}$$
$$U_3(1,0,z)=-\frac{1}{4\pi(1+z^2)^{\frac{1}{2}}}$$
$$U_2=2[\int_0 ^N U_3(x,y,z)-\int_0 ^N U_3(1,0,z)]dz$$
$$\implies$$
$$U_2=-\frac{1}{2\pi}[\int_0 ^N\frac{1}{\sqrt{x^2+y^2+z^2}}+\int _0 ^N \frac{1}{\sqrt{1+z^2}}]$$
$$=-\frac{1}{2\pi}[\log (z+\sqrt{x^2+y^2+z^2})\Big {|}_0 ^N-\log (z+\sqrt{1+z^2})\Big{|}_0 ^N]$$
$$U_2=-\frac{1}{2\pi}[\log \frac{N+\sqrt{x^2+y^2+z^2}}{N+\sqrt{1+N^2}}]-\log \sqrt{x^2+y^2}$$
$$\text{As} N\implies \infty$$
$$ \log \frac{N+\sqrt{x^2+y^2+z^2}}{N+\sqrt{1+N^2}}]\implies 0$$
$$\text{Therefore we have}$$
$$U_2=\frac{1}{2\pi} \log \sqrt{x^2+y^2}   \blacksquare$$

[Victor Ivrii]

Don't use $\implies$ instead of $\to$  (etc) as the former is a logical sign