### Author Topic: Exam Week  (Read 1111 times)

#### Victor Ivrii ##### Exam Week
« 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 $w=u-v$ will satisfy (\ref{1}), homogeneous condition (\ref{2}), modified (\ref{3}). Find $w$ by separation, and then $u=v+w$.

#### Andrew Hardy

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• Karma: 10 ##### Re: Exam Week
« Reply #1 on: April 12, 2018, 10:17:00 PM »
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) ?
« Last Edit: April 13, 2018, 08:54:11 AM by Andrew Hardy »

#### Victor Ivrii ##### Re: Exam Week
« Reply #2 on: April 13, 2018, 12:21:39 AM »
Quote
How do I modify (3) ?
Calculate $w|_{x^2+y^2=1}$ as $y>0$, if $u$ satisfies (\ref{3}) and $v=x^2-y^2$ you know.

#### Andrew Hardy

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• Karma: 10 ##### Re: Exam Week
« Reply #3 on: April 13, 2018, 09:08:11 AM »
To calculate $w|_{x^2+y^2=1}$ I make a substitution that if  $$u|_{x^2+y^2 =1}=1$$ then  $$v|_{x^2+y^2 =1}= 1 - y^2 - y^2 = 1-2y^2$$
$$v|_{x^2+y^2 =1}= x^2 - (1-x^2) = 2x^2 - 1$$
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.

#### Sheng Gao

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• • Posts: 3
• Karma: 0 ##### Re: Exam Week
« Reply #4 on: April 13, 2018, 11:48:40 PM »
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 $w=u-v, then \Delta w=0$ $w|_{y=0}=0$, $w|_{x^2+y^2=1}=2sin^2\theta$

#### Victor Ivrii ##### Re: Exam Week
« Reply #5 on: April 14, 2018, 05:07:22 AM »
Sheng
correct, but one needs to solve the problem for $w$ by the standard separation of variables.

#### Andrew Hardy

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• Karma: 10 ##### Re: Exam Week
« Reply #6 on: April 14, 2018, 02:12:26 PM »
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
$$u = r^2 \cos(2 \theta) + \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta)$$