### Author Topic: FE-P4  (Read 1197 times)

#### Victor Ivrii

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##### FE-P4
« on: April 11, 2018, 02:41:32 PM »
Consider the Laplace equation in the sector
\begin{align}
&u_{xx} +  u_{yy} =0\qquad &&\text{in    } \frac{1}{4} \le x^2+y^2 < 4, y>0  ,
\label{4-1} \\
& u =1\qquad &&\text{for  }  x^2+y^2=4,\label{4-2}\\
& u =-1\qquad &&\text{for  }  x^2+y^2=\frac{1}{4},\label{4-3}\\
&  u=0 &&\text{for\ \ }  y=0 ,\label{4-4}
\end{align}
where $\theta$ is a polar angle.

a  Look for solutions $u$ in the form of  $u(r,\theta)= R(r) P(\theta)$  (in polar coordinates) and derive a set of
ordinary differential equations for $R$ and $P$. Write the correct  boundary conditions for $P$.

b  Solve the eigenvalue problem for $P$ and find all eigenvalues.

c  Solve the differential equation  for $R$.

d  Find the solution $u$ of (\ref{4-1})--(\ref{4-4}).
« Last Edit: April 13, 2018, 08:24:54 AM by Victor Ivrii »

#### Andrew Hardy

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##### Re: FE-P4
« Reply #1 on: April 11, 2018, 05:55:02 PM »
a) separation of variables gives us that $$P'' + \lambda P = 0$$
$$r^2R''+ rR' -\lambda R = 0$$
The boundary conditions translate to Dirichlet Boundary Conditions $P(0) = P(\theta) = 0$
b) With the eigenvalue problem of  $$P \lambda = (n)^2$$ gives $$P = \sin(n\theta)$$
c) and then solving the ODE for R we know $$R = Ar^n + Br^{-n}$$
d) Now solving for coefficients $$u = \sin(n\theta)(Ar^n + Br^{-n} )$$
Now we know $A = -B$  because  of (2),(3) and so
$$A_n =\frac{2}{\pi} \int_{0}^{\pi} \frac{1}{2^{-n}-2^{n}} \sin(n\theta)$$ and via computation
becomes $$u =\sum_{n \text{ odd}} \frac{4}{n\pi(2^n-2^{-n})} \sin(n\theta) ( r^n + r^{-n})$$
« Last Edit: April 11, 2018, 06:11:00 PM by Andrew Hardy »

#### George Lu

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##### Re: FE-P4
« Reply #2 on: April 11, 2018, 06:07:34 PM »
First, we convert the Laplacian to polar coordinates: $\Delta u =u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}$. Write $u=RP$, so that we can solve the problem of form $\frac{r^2R''+rR'}{R}+\frac{P''}{P}=0$

Now, notice that boundary condition (4) becomes $u(r,0)=u(r,\pi)=0$ in polar coordinates. This gives us a condition for periodicity in P, so that we can associate a negative eigenvalue $-\lambda_n=-\omega_n^2$ to it. Solving $P''+\omega_n^2 P=0$ gives us $P=A_n \cos (\omega_n \theta) + B_n \sin (\omega_n \theta)$. We can immediately eliminate the $A_n$ term, because of the boundary condition on $\theta=0$, and we can also conclude that $\omega_n \pi = n \pi$ means that our $\omega_n=n$, $\lambda = n^2$.

Now, we expect some equations of the form $r^m$ for the solution to R. Solving the Euler problem $r^2R+rR'=\lambda R$ gives $m(m-1)+m=n^2\rightarrow m = \pm n$. So $R=A_n r^n+B_n r^{-n}$. Neither term can be negated, as r does not tend to 0 or infinity. However, $A_0$ and $B_0$ can be ignored as they do not obey the periodic behaviour we expect.

This gives general form $u(r,\theta) = \sum_{n=1}^\infty(A_n r^n+B_n r^{-n})\sin (n \theta)$. Now, we consider boundary conditions (2) and (3) in the cases $r=2, r=2^{-1}$ respectively.

Then

$u(2,\theta) = \sum_{n=1}^\infty(A_n 2^n+B_n 2^{-n})\sin (n \theta) = 1$ (2)

$u(2^{-1},\theta) = \sum_{n=1}^\infty(A_n 2^{-n}+B_n 2^{n})\sin (n \theta) = -1$ (3)

If we add these two summations together, this implies that $B_n=-A_n$. Use this simplification to find the fourier coefficients in (2):

$u(2,\theta) = \sum_{n=1}^\infty A_n (2^n - 2^{-n})\sin (n \theta) = 1$

Substitute $C_n=A_n(2^n-2^{-n})$. Then $C_n=\frac{2}{\pi}\int_0^\pi sin (n \theta) d\theta = \frac{4}{n\pi}$ for n odd, $0$ for n even.

So $u(r,\theta) = \sum_{n=1}^\infty \frac{4}{n\pi(2^n-2^{-n})} (r^n - r^{-n})\sin (n \theta)$

#### Victor Ivrii

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##### Re: FE-P4
« Reply #3 on: April 13, 2018, 06:40:51 AM »
Andrew
Missing calculations for coefficients

George
You proved that $A_n=0$ for even $n$, but in the final formula you sum over all $n$. [/color]

Solution
In polar coordinates $\{y=0\}$ is $\theta =0,\pi$.

Separating variables we get
\begin{align}
\frac{r^2 R''+rR'}{R}+\frac{P''}{P}=0\implies  &P''+\lambda P=0,\label{4-5}\\
&P(0)=P(\pi)=0,\label{4-6}\\
&r^2R''+rR'+\lambda R=0.\label{4-7}
\end{align}
This problem has solutions
$\lambda_n =n^2$, $X_n=\sin (n\theta)$, $n=1,2,\ldots$.

Then $r^2R''+rR'+n^2 R=0\implies R_n= A_n r^{n} + B_n r^{-n}$.  Then
\begin{align}
&u= \sum_{n=1}^\infty \bigl(A_n r^{n} +B_{n}r^{-n}\bigr) \sin (n\theta)
\label{G}
\end{align}
and using (\ref{4-2}), (\ref{4-3})
\begin{align*}
&u|_{r=2}= \sum_{n=1}^\infty (2^{n}A_n +2^{-n}B_n)\sin(n\theta)=1,\\
&u|_{r=\frac{1}{2}}= \sum_{n=1}^\infty (2^{-n}A_n +2^{n}B_n)\sin(n\theta)=-1\\
\end{align*}
which imply
\begin{align*}
&\sum_{n=1}^\infty (2^n+2^{-n})(A_n+B_n)\sin(n\theta)=0\implies (A_n+B_n)=0\\
&\sum_{n=1}^\infty (2^n-2^{-n})(A_n-B_n)\sin(n\theta)=2\implies
(2^n-2^{-n})(A_n-B_n)=\\
&\frac{4}{\pi}\int_0^\pi\sin(n\theta)\,d\theta=
\left\{\begin{aligned}
&\frac{8}{\pi (2m+1)}&& n=2m+1,\\
&0 &&n=2m\
\end{aligned}\right.
\end{align*}
which imply  $A_{2m}=B_{2m}=0$ and
\begin{align*}
&A_{2m+1}=-B_{2m+1}= \frac{4}{\pi (2m+1)(2^{2m+1}-2^{-2m-1})}
\end{align*}
Finally
\begin{align*}
u= &\sum_{m=0}^\infty \frac{4}{\pi (2m+1)(2^{2m+1}-2^{-2m-1})}(r^{2m+1}-r^{-2m-1})
\sin ((2m+1)\theta).
\end{align*}

* Wrong bounds for $\theta$ and wrong b.c. for $P$ (usually periodic b.c., rather then $P(0)=P(\pi)=0$; then wrong eigenfunctions.
* Wrong bonds for $r$: $\frac{1}{4} \le r\le 4$ rather than $\frac{1}{2} \le r\le 2$.
* Throwing out $r^{-n}$ because it is unbounded as $r\to 0$ (but $r\ge 1/2$).
* Instead of plugging  (\ref{G}) into (\ref{4-2}), (\ref{4-3}) something like plugging with $r^n$ into (\ref{4-2}) and with $r^{-n}$ into (\ref{4-3})
* Wrong solutions to Euler's equation. $R_n(r)=e^{nr}$ and $R_n(r)=e^{-nr}$ (and variants) rather than $R_n(r)=r^n$ and $R_n(r)=r^{-n}$.
* Again, separate solutions for odd and even $n$.