Author Topic: TT2--Problem 5  (Read 4709 times)

Victor Ivrii

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TT2--Problem 5
« 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*}

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« Last Edit: November 15, 2012, 09:38:01 PM by Victor Ivrii »

Ian Kivlichan

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Re: TT2--Problem 5
« Reply #1 on: November 15, 2012, 10:31:07 PM »
Hopeful solution attached! :)

Chen Ge Qu

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Re: TT2--Problem 5
« Reply #2 on: November 15, 2012, 10:32:17 PM »
See attached

Victor Ivrii

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Re: TT2--Problem 5
« Reply #3 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.
\qquad\qquad\qquad
w = \left\{\begin{aligned}
&y &&\text{as  }x=1,\\[3pt]
-&y &&\text{as }x=-1,\\[3pt]
&0 &&\text{as   }y=1,\\[3pt]
&0 &&\text{as   }y=-1.
\end{aligned}\right.
\end{equation*}
and then separate variables and after looooong calculations get solutions in the form of series.

But $u=xy$ is correct and simple (much simpler than formulae for $v$, $w$).

Sometimes insight (or just luck) beats an expertise!
« Last Edit: November 16, 2012, 07:11:56 AM by Victor Ivrii »