find the equation for outer side of circle using Neumann method.
$\Delta u= F(r,\varphi),$
$ 0<r_0<r, $
$0\leq \varphi<2\pi,$
$u_{r}(r_0,\varphi = f(\varphi), \left| u \right|<\infty),$
if,
$ F= \frac{A\ln r}{r^3}\times \cos2\varphi$
solution:
$\Delta u= \frac{A\ln r}{r^3} \times \cos 2\varphi $
$\frac{1}{r} \frac{\partial }{\partial x} r \frac{\partial u}{\partial x}+\frac{1}{r^2}\frac{\partial^2 \varphi}{\partial x^2} =\frac{A\ln r}{r^3}\times\cos2\varphi $
$ \frac{\partial^2 u}{\partial x^2}+ \frac{1}{r}\frac{\partial r }{\partial x} \frac{\partial u }{\partial x} = \frac{A\ln r}{r^3}\times\cos2\varphi $
let,
$u(r,\varphi)=R(r)\Theta(\theta)$
$R''\Theta+\frac{1}{r}R'\Theta+\frac{1}{r^2}R\Theta''=0;$
$r^2R''+rR'-\lambda R=0..............................(1)$
$\Theta''+\lambda\Theta=0...........................................(2)$
I used cases for $\lambda$:
- $\lambda < 0 $
say,
$ \lambda = -\mu ^2( \mu > 0)$
so
$\Theta '' - \mu ^2\Theta = 0$
$\Theta(\theta)=C_1 e^{\mu \theta} + C_2 e^{-\mu \theta},$
since $\theta $ is periodic $C_1=C_2=0$
for case 2:
$\lambda=0; \Theta ''=0$
$\Theta(\theta)=A\theta+B$
since $\theta $ is periodic $A=0$, $\Theta(\theta)= B $ or just 1
$\lambda = 0$
$r^2R''+rR'=0$
using Eular theorem,
$s(s-1)+s=0$
$s^2=0$
$s=0,0$
$R(r)=C_1r^0+C_2r^0\ln r= C_1+C_2\ln r$
for $r\rightarrow 0$,
$u(r,\theta)= R(r)\Theta(\theta)=1$
case 3:
$\lambda>0$
say $\lambda = \mu ^2 (\mu>0)$
so i found $\Theta(\theta) = C_1\cos(\mu \theta)+C_2\sin(\mu \theta)$
with euler theorem
$r^2R''+rR'-n^2R=0$
$s(s-1)+s-n^2=0$
$s=\pm n$
$R(r)=C_1r^n+C_2r^{-n}$
if r\rightarrow 0, C_2=0;
$R(r)= C_1r^n = Ar^n$
thus $u(r,\theta)= Ar^n(C_1cos(n\theta)- C_2sin(n\theta))$
now i don't know how to do it.
please can anyone complete it for me.