# Toronto Math Forum

## APM346-2012 => APM346 Math => Term Test 2 => Topic started by: Victor Ivrii on November 15, 2012, 08:26:34 PM

Title: TT2--Problem 5
Post by: Victor Ivrii on November 15, 2012, 08:26:34 PM
Let $Q = \{(x,y)\in {\mathbb{R}}^2: |x|<1, |y|<1\}.$ Draw the set $Q.$ We define data $g$ on the boundary of $Q$:

Find the solution $u$ of the Dirichlet problem on $Q$:
\begin{equation*}
\Delta u=0 \qquad \text{for  }  (x,y) \in Q
\end{equation*}
with the boundary conditions
\begin{equation*}
u = \left\{\begin{aligned}
&y &&\text{as  }x=1,\\[3pt]
-&y &&\text{as }x=-1,\\[3pt]
&x &&\text{as   }y=1,\\[3pt]
-&x &&\text{as   }y=-1.
\end{aligned}\right.
\end{equation*}

Post after 22:30
Title: Re: TT2--Problem 5
Post by: Ian Kivlichan on November 15, 2012, 10:31:07 PM
Hopeful solution attached! :)
Title: Re: TT2--Problem 5
Post by: Chen Ge Qu on November 15, 2012, 10:32:17 PM
See attached
Title: Re: TT2--Problem 5
Post by: Victor Ivrii on November 16, 2012, 07:10:05 AM
Actually method of separation should not be applied in the case when all the boundary conditions are inhomogeneous and an idiot-expert would write $u=v+w$ where $v$, $w$ satisfy the same equation but with
\begin{equation*}
v = \left\{\begin{aligned}
&0 &&\text{as  }x=1,\\[3pt]
&0 &&\text{as }x=-1,\\[3pt]
&x &&\text{as   }y=1,\\[3pt]
-&x &&\text{as   }y=-1,
\end{aligned}\right.
But $u=xy$ is correct and simple (much simpler than formulae for $v$, $w$).