### Author Topic: FE4  (Read 3327 times)

#### Victor Ivrii ##### FE4
« on: December 13, 2016, 07:55:55 PM »
Consider the Laplace equation in the sector
\begin{align}
&u_{xx} +  u_{yy} =0\qquad &&\text{in  } x^2+y^2 < 16, x> -\sqrt{3}|y|  ,
\label{4-1}
\end{align}
with the boundary conditions
\begin{align}
& u =1\qquad &&\text{for  }  x^2+y^2=16,\label{4-2}\\
& u=0 &&\text{for  }  x=-\sqrt{3}|y| .\label{4-3}
\end{align}

• 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 $P$.
• Solve the eigenvalue problem for $P$ and find all eigenvalues.
• Solve the differential equation  for $R$.
• Find the solution $u$ of (\ref{4-1})--(\ref{4-3}).

« Last Edit: December 13, 2016, 08:11:59 PM by Victor Ivrii »

#### Luyu CEN

• Jr. Member
•  • Posts: 12
• Karma: 0 ##### Re: FE4
« Reply #1 on: December 14, 2016, 12:58:27 PM »
\begin{equation}
\Delta =\partial_r^2 + \frac{1}{r}\partial_r +
\frac{1}{r^2}\partial_\theta^2
\end{equation}
Let's plug in $u = R(r)P(\theta)$ and I get
\begin{equation*}
R''(r)P(\theta) + \frac{1}{r}R' (r)P(\theta) + \frac{1}{r^2} R(r) P''(\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{r^2 R''(\rho) + r R' (r)}{R(r)}+
\frac{P''(\theta)}{P(\theta)}=0
\end{equation*}
The two terms must be constant
\begin{align}
&r^2 R'' +r R' = \lambda R\\[3pt]
&P(\theta)=-\lambda P(\theta).
\end{align}
The initial conditions can be translated as
\begin{align}
& u =1\qquad &&\text{for  }  r = 4,\\
& u=0 &&\text{for  }  \theta = -\pi/6, 5\pi/6 \end{align}
Therefore the boundary condition for P is $P(-\frac{\pi}{6}) = P(\frac{5\pi}{6}) = 0$ and this is a Dirichlet-Dirichlet boundary problem with $l = \pi$.
Thus we have
\begin{align}
&\lambda_{n} = n^2
&P_{n} = sin(n(\theta + \frac{\pi}{6}))
&& n = 1, 2, ...
\end{align}
\begin{equation}
r^2 R'' +r R' - n^2 R = 0
\end{equation}
The Euler equation for this is $k(k-1) + k - n^2 = 0$ and we have $k = n, -n$
Since we are looking for continuous solutions,
\begin{equation}
R_n = A_n r^n
u = \sum_{n = 1}^{\infty}A_n r^n sin(n(\theta + \frac{\pi}{6}))
\end{equation}
Apply the initial condition
\begin{equation}
\sum_{n = 1}^{\infty}A_n 4^n sin(n(\theta + \frac{\pi}{6})) = 1
\end{equation}
Then we calculate the coefficients by
\begin{equation}
A_n=\frac{2}{4^n \pi}\int_{-\frac{\pi}{6}}^{\frac{5\pi}{6}}\sin(n(\theta+\frac{\pi}{6}))\,d\theta=\frac{2}{4^n \pi}\int_0^{\pi}\sin(nx) dx=\frac{2}{n4^n \pi}(\cos(nx)|_{\pi}^0)=\frac{1}{n4^{n-1} \pi}\,\,\,\text{n odd, 0 otherwise}
\end{equation}
Therefore
\begin{equation}
u = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{1}{n4^{n-1} \pi}r^n sin(n(\theta + \frac{\pi}{6}))
\end{equation}
« Last Edit: December 14, 2016, 01:00:35 PM by Luyu CEN »

#### XinYu Zheng

• Full Member
•   • Posts: 27
• Karma: 0 ##### Re: FE4
« Reply #2 on: December 14, 2016, 03:34:45 PM »
Note: The sector is not a half circle. There is an absolute value on $y$.

#### Luyu CEN

• Jr. Member
•  • Posts: 12
• Karma: 0 ##### Re: FE4
« Reply #3 on: December 14, 2016, 04:56:26 PM »
Yes thanks. I modify my solution to solve for the sector instead of the half disk
\begin{equation}
\Delta =\partial_r^2 + \frac{1}{r}\partial_r +
\frac{1}{r^2}\partial_\theta^2
\end{equation}
Let's plug in $u = R(r)P(\theta)$ and I get
\begin{equation*}
R''(r)P(\theta) + \frac{1}{r}R' (r)P(\theta) + \frac{1}{r^2} R(r) P''(\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{r^2 R''(\rho) + r R' (r)}{R(r)}+
\frac{P''(\theta)}{P(\theta)}=0
\end{equation*}
The two terms must be constant
\begin{align}
&r^2 R'' +r R' = \lambda R\\[3pt]
&P(\theta)=-\lambda P(\theta).
\end{align}
The initial conditions can be translated as
\begin{align}
& u =1\qquad &&\text{for  }  r = 4,\\
& u=0 &&\text{for  }  \theta = -\frac{5\pi}{6}, \frac{5\pi}{6} \end{align}
Therefore the boundary condition for P is $P(-\frac{5\pi}{6}) = P(\frac{5\pi}{6}) = 0$ and this is a Dirichlet-Dirichlet boundary problem with $l = \frac{5\pi}{3}$.
Thus we have
\begin{align}
&\lambda_{n} = (\frac{3n}{5})^2
&P_{n} = \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
&& n = 1, 2, ...
\end{align}
\begin{equation}
r^2 R'' +r R' - (\frac{3n}{5})^2 R = 0
\end{equation}
The Euler equation for this is $k(k-1) + k - (\frac{3n}{5})^2 = 0$ and we have $k = \frac{3n}{5}, -\frac{3n}{5}$
Since we are looking for continuous solutions,
\begin{equation}
R_n = A_n r^{\frac{3n}{5}}\\
u = \sum_{n = 1}^{\infty}A_n r^\frac{3n}{5} \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
\end{equation}
Apply the initial condition
\begin{equation}
\sum_{n = 1}^{\infty}A_n 4^\frac{3n}{5} \sin(n(\theta + \frac{5\pi}{6})) = 1
\end{equation}
Then we calculate the coefficients by
\begin{equation}
A_n=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}}\int_{-\frac{5\pi}{6}}^{\frac{5\pi}{6}}\sin(\frac{3n}{5}(\theta+\frac{5\pi}{6}))\,d\theta=\frac{2}{4^\frac{3n}{5} \frac{5\pi}{3}}\int_0^{\frac{5\pi}{3}}\sin(\frac{3n}{5}x) \,dx=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}\frac{3n}{5}}(\cos(\frac{3n}{5}x)|_{\frac{5\pi}{3}}^0)=\frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}\,\,\,\text{n odd, 0 otherwise}
\end{equation}
Therefore
\begin{equation}
u = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6})) = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{100}{9n\pi4^{\frac{3n}{5}}} r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6}))
\end{equation}
« Last Edit: December 19, 2016, 10:14:47 AM by Luyu CEN »

#### Victor Ivrii ##### Re: FE4
« Reply #4 on: December 17, 2016, 11:03:05 AM »
Correct. Can you simplify the answer?

Also, one can simplify calculations observing that the problem is symmetric with respect to $y=0$ (that means $\theta=0$), in particular, because $U|_{r=4}=1$ is an even function. Therefore from the very beginning we need to consider $\cos(\omega \theta)$, s.t. $\cos(\omega \times 4\pi/6)=0$, that is $5\pi\omega /6= (2m+1)\pi /2\implies \omega=\omega_m= =3(2m+1) /5$, $\lambda_m= 9(2m+1)/25$.

Many students thought of $2\pi$-periodic $P(\theta)$ which is not the case as a membrane is not disk, but a sector.
« Last Edit: December 17, 2016, 12:32:33 PM by Victor Ivrii »

#### Victor Ivrii ##### Re: FE4
« Reply #5 on: December 17, 2016, 12:44:42 PM »
Further, even if the shape was a disk, three cases are different:

(1) Circular membrane; then $P''=-\lambda P$ for all $\theta$, $P$ is $2\pi$-periodic;
(2) Circular membrane, with a  cut along $\theta=\pi$,  and it is fixed there; then $P''=-\lambda P$ for  $-\pi<\theta<\pi$, $P(-\pi)=P(\pi)=0$;
(3) Circular membrane, with a  cut along $\theta=\pi$,  and it is free there; then $P''=-\lambda P$ for  $-\pi<\theta<\pi$, $P'(-\pi)=P'(\pi)=0$.