Author Topic: TT2-P4  (Read 213 times)

Victor Ivrii

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TT2-P4
« on: November 17, 2016, 03:27:07 AM »
Consider Laplace equation in the sector
\begin{equation}
u_{rr} +\frac{1}{r}u_r +\frac{1}{r^2}u_{\theta\theta}=0 \qquad  r<8,\,0<\theta<\frac{3}{2}\pi \label{4-1}
\end{equation}
with the Dirichlet boundary conditions as $\theta=0$ and $\theta=\frac{3}{2}\pi$
\begin{equation}
u|_{\theta=0}=u|_{\theta=\frac{3}{2}\pi}=0\label{4-2}\\
\end{equation}
and the Dirichlet boundary condition as $r=8$
\begin{equation}
u|_{r=8}=1.\label{4-3}
\end{equation}
Using separation of variables find solution as a series.

XinYu Zheng

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Re: TT2-P4
« Reply #1 on: November 17, 2016, 08:08:36 AM »
Introduce $u=R(r)\Theta(\theta)$. Then after separating variables the angular equation will be $\Theta''+\lambda\Theta=0$ with Dirichlet B.C at 0 and $3\pi/2$. We know the solution to this problem:
$$\Theta_n=\sin(2n\theta/3)$$
With $\lambda_n=4n^2/9$, $n\geq 1$. Then the radial equation will have solution $R(r)=Ar^{2n/3}+Br^{-2n/3}$. We will have to drop the second term because they blow up at the origin. Thus our general solution is
$$u(r,\theta)=\sum_{n\geq 1} A_nr^{2n/3}\sin(2n\theta/3)$$
Now applying the B.C. at $r=8$ we have
$$1=\sum_{n\geq 1} A_n4^n\sin(2n\theta/3)$$
And the coefficients can be calculated the usual way:
$$A_n=\frac{1}{4^n}\frac{2}{\frac{3\pi}{2}}\int_0^{3\pi/2}\sin(2n\theta/3)\,d\theta=\frac{1}{4^n}\frac{4}{3\pi}\frac{3}{2n}\cos(2n\theta/3)|_{3\pi/2}^0=\frac{1}{4^{n-1}}\frac{1}{2n\pi}(1-(-1)^n)=\frac{1}{4^{n-1}}\frac{1}{n\pi}\,\,\,\,\,\,\text{n odd, 0 otherwise}$$
So our solution is
$$u(r,\theta)=\sum_{n\geq 1, n\,\,odd}\frac{1}{4^{n-1}}\frac{1}{n\pi}r^{2n/3}\sin(2n\theta/3)$$