Toronto Math Forum
APM3462018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on April 12, 2018, 03:38:18 PM

Consider problem:
\begin{align}
& \Delta u=0 &&\text{in }x^2+y^2<1, \ \ y>0,
\label{1}\\
& u_{y=0}=x^2 &&\text{as }x<1,\label{2}\\
& u_{x^2+y^2=1}=1,&& \text{as } y>0.
\label{3}
\end{align}
We want to separate variable $r$ and $\theta$ but the conditions as $\theta=0,\pi$ are inhomogeneous.
So we want to make them homogeneous. Find $v$, so that $u:=v$ satisfies (\ref{1}) and (\ref{2}) but not necessarily (\ref{3}), so $v$ is not unique. Can you suggest a candidate?
Then $w=uv$ will satisfy (\ref{1}), homogeneous condition (\ref{2}), modified (\ref{3}). Find $w$ by separation, and then $u=v+w$.

Only considering (1) and (2),
My candidate is
$$ v = x^2 y^2 $$
which is harmonic, (has zero laplacian) and $$ v(x,0) = x^2 $$
By homogeneous (2) do you mean?
$$ w_{y=0}=0 $$
How do I modify (3) ?

How do I modify (3) ?
Calculate $w_{x^2+y^2=1}$ as $y>0$, if $u$ satisfies (\ref{3}) and $v=x^2y^2$ you know.

To calculate $w_{x^2+y^2=1} $ I make a substitution that if $$u_{x^2+y^2 =1}=1 $$ then $$ v_{x^2+y^2 =1}= 1  y^2  y^2 = 12y^2 $$
$$ v_{x^2+y^2 =1}= x^2  (1x^2) = 2x^2  1 $$
so then $$w_{x^2+y^2=1}=2y^2 = 2 2x^2$$
Not sure if I'm on the right track, I'll return to this later today.

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 $v=x^2y^2$, then
$\Delta v=v_{xx}+v_{yy}=22=0$
$v_{y=0}=x^2$ $v_{x^2+y^2=1}=12y^2=12r^2sin\theta=12sin\theta, since$ $r=1$
And
Let $w=uv, then \Delta w=0$ $w_{y=0}=0$, $w_{x^2+y^2=1}=2sin^2\theta$

Sheng
correct, but one needs to solve the problem for $w$ by the standard separation of variables.
Please, escape \sin , \cos

Writing the change of coordinates Sheng applied to the boundary conditions I found, $ w_{\pi=0}=0$ and $ w_{r=1}=2\sin^2\theta $ I can apply separation of variables to the function $ w = P(\theta)R(r) $ which I can solve for this half disk as
$$ w = \sum A r^n \sin(n\theta) $$ and therefore
$$ A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta =\frac{ (4 (2 \cos(π n)  2))}{(π (n^3  4 n)) }$$
and this is $ \frac{8}{(π (n^3  4 n)) } $ when n is odd. So I have
$$ w = \sum_{n \text{ odd}} \frac{8}{(π (n^3  4 n)) } r^n \sin(n\theta) $$
and therefore I can write $$ u = w + v = \sum_{n \text{ odd}} \frac{8}{(π (n^3  4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta)  r^2 sin^2(\theta) $$ which can be written most concisely as
$$ u = r^2 \cos(2 \theta) + \sum_{n \text{ odd}} \frac{8}{(π (n^3  4 n)) } r^n \sin(n\theta) $$