Toronto Math Forum

By Laplace equation
$u_{rr} +\frac{1}{r}u_r +\frac{1}{r^2}u_{\theta\theta}=0,  r<1, 0<\theta<\pi$
$u(r,0)=r^2$
$u(r,\pi)=r^2$
$u|_{r=1}=1$

let $v=x^2-y^2$, then
$\Delta v=v_{xx}+v_{yy}=2-2=0$
$v|_{y=0}=x^2$  $v|_{x^2+y^2=1}=1-2y^2=1-2r^2sin\theta=1-2sin\theta, since$  $r=1$

And
Let $w=u-v, then    \Delta w=0$ $w|_{y=0}=0$, $w|_{x^2+y^2=1}=2sin^2\theta$
                                         

$u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ where $\phi$, $\psi$ so far are unknown and will be chosen later. Such $u$ obviously exist for any $\phi$, $\psi$.

What about $v:=u_{yy}-g=0$? Obviously $v_{xx}=0$ and we need to satisfy $v(0,y)=0$, $v_x(0,y)=0$, which are, in fact, $\phi '' -g(0,y)=0$, $\psi ''- g_x(0,y)=0$, and we can satisfy these two equations.  Four arbitrary constants $C_1=\phi(0)$, $C_2=\phi'(0)$, $C_3=\psi(0)$ and $C_4=\psi'(0)$ appear.

May I ask ask about how we get $u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$? Since I cannot follow up starting here.....