Show Posts
1
Quiz-6 / Quiz 6 T0101
« on: March 16, 2018, 01:44:27 AM »
Solve
\begin{align*}
& \Delta u:=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt]
& u_r|_{r=a}=f(\theta).
\end{align*}
where we use polar coordinates $(r,\theta)$ and $f(\theta)=\left\{\begin{aligned}
&1 &&0<\theta<\pi,\\
-&1 &&\pi<\theta<2\pi.
\end{aligned}\right.$
The expected answer: solution as a series.
Given the condition $r<a$. We can discard the logr and $r^{-n}$ part ,so the general solution is
\begin{gather*}
u(r,\theta) =\frac{1}{2}A_0+\sum_{n=1}^\infty r^{n} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(r,\theta)=\sum_{n=1}^\infty nr^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(a,\theta)=\sum_{n=1}^\infty na^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta)) = f(\theta).
\end{gather*}
For $n = 0,1,2,3\ldots$
$$ A_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\cos}(n\theta)d\theta $$
For $n = 1,2,3\ldots$
$$ B_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\sin}(n\theta)d\theta $$
Since $f(\theta)$ is odd function $\implies$ $A_n$ = 0 and
\begin{gather*}B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi} f(\theta){\sin}(n\theta)d\theta \\
\implies B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi}{\sin}(n\theta)d\theta
\end{gather*}
We integral this get $-\frac{1}{n}\cos(n\theta)$ from 0 to $\pi$
$$\implies B_n = \frac{2}{{\pi}n^{2}a^{n-1}}[1-(-1)^{n}] $$
If we write n = 2m, it becomes 0, so we consider n = 2m+1
\begin{gather*}\implies B_n = \frac{4}{{\pi}(2m+1)^{2}a^{2m}}\\
u(r,\theta) = \frac{4}{{\pi}}\sum_{m=0}^\infty \frac{r^{2m+1}}{(2m+1)^{2}a^{2m}} \sin(2m+1)\theta
\end{gather*}
\begin{align*}
& \Delta u:=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt]
& u_r|_{r=a}=f(\theta).
\end{align*}
where we use polar coordinates $(r,\theta)$ and $f(\theta)=\left\{\begin{aligned}
&1 &&0<\theta<\pi,\\
-&1 &&\pi<\theta<2\pi.
\end{aligned}\right.$
The expected answer: solution as a series.
Given the condition $r<a$. We can discard the logr and $r^{-n}$ part ,so the general solution is
\begin{gather*}
u(r,\theta) =\frac{1}{2}A_0+\sum_{n=1}^\infty r^{n} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(r,\theta)=\sum_{n=1}^\infty nr^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(a,\theta)=\sum_{n=1}^\infty na^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta)) = f(\theta).
\end{gather*}
For $n = 0,1,2,3\ldots$
$$ A_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\cos}(n\theta)d\theta $$
For $n = 1,2,3\ldots$
$$ B_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\sin}(n\theta)d\theta $$
Since $f(\theta)$ is odd function $\implies$ $A_n$ = 0 and
\begin{gather*}B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi} f(\theta){\sin}(n\theta)d\theta \\
\implies B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi}{\sin}(n\theta)d\theta
\end{gather*}
We integral this get $-\frac{1}{n}\cos(n\theta)$ from 0 to $\pi$
$$\implies B_n = \frac{2}{{\pi}n^{2}a^{n-1}}[1-(-1)^{n}] $$
If we write n = 2m, it becomes 0, so we consider n = 2m+1
\begin{gather*}\implies B_n = \frac{4}{{\pi}(2m+1)^{2}a^{2m}}\\
u(r,\theta) = \frac{4}{{\pi}}\sum_{m=0}^\infty \frac{r^{2m+1}}{(2m+1)^{2}a^{2m}} \sin(2m+1)\theta
\end{gather*}